The vector a = αi + 2j + βk lies in the plane of the vectors b = i + j and c = j + k and bisects the

Q: Why can't we just use the cross product to find the plane?

The cross product $\vec{b} \times \vec{c}$ gives the normal to the plane, not the bisector. While it defines the plane, it doesn't help us find the direction of the bisector within that plane. We need the bisector itself, not the normal.

Q: Wait, why do we assume $\vec{a} = \lambda(\vec{b} + \vec{c})$? What if $\vec{a}$ is in the opposite direction?

The vector $\vec{a}$ is an internal angle bisector, which means it must point in the same general direction as the sum of the vectors. The scalar $\lambda$ accounts for the magnitude of $\vec{a}$. If $\lambda$ were negative, it would point in the opposite direction, which would be the external bisector, not the internal one.

Q: Do we always need to check the magnitudes? What if $|\vec{b}| \neq |\vec{c}|$?

Excellent question! If the magnitudes were different, the sum $\vec{b} + \vec{c}$ would not be the bisector. You would have to normalize them first: $\hat{b} + \hat{c}$. Always check the magnitudes first—it's a crucial step that can save you from a lot of unnecessary algebra.

Question

The vector $\vec{a} = \alpha\hat{i} + 2\hat{j} + \beta\hat{k}$ lies in the plane of the vectors $\vec{b} = \hat{i} + \hat{j}$ and $\vec{c} = \hat{j} + \hat{k}$ and bisects the angle between $\vec{b}$ and $\vec{c}$. Then which one of the following gives possible values of $\alpha$ and $\beta$?

Options:

(A) $\alpha = 2, \beta = 2$

(B) $\alpha = 1, \beta = 2$

(C) $\alpha = 2, \beta = 1$

(D) $\alpha = 1, \beta = 1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two reference vectors, $\vec{b} = \hat{i} + \hat{j}$ and $\vec{c} = \hat{j} + \hat{k}$. These vectors define a plane in 3D space.

A third vector, $\vec{a} = \alpha\hat{i} + 2\hat{j} + \beta\hat{k}$, lies within this plane and acts as the angle bisector between $\vec{b}$ and $\vec{c}$.

### The Secret of Equal Magnitudes

The internal angle bisector of two vectors $\vec{x}$ and $\vec{y}$ is generally directed along the sum of their unit vectors:
$$\frac{\vec{x}}{|\vec{x}|} + \frac{\vec{y}}{|\vec{y}|}$$

However, if the two vectors have the same magnitude, their simple sum $\vec{x} + \vec{y}$ points exactly along the angle bisector. Let us calculate the magnitudes of our given vectors:

$$|\vec{b}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}$$

$$|\vec{c}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$$

Since $|\vec{b}| = |\vec{c}| = \sqrt{2}$, the bisector is parallel to the resultant vector $\vec{d} = \vec{b} + \vec{c}$.

### The Algebraic Bridge

We construct the resultant vector $\vec{d}$ by adding the components of $\vec{b}$ and $\vec{c}$:

$$\vec{d} = (\hat{i} + \hat{j}) + (\hat{j} + \hat{k}) = \hat{i} + 2\hat{j} + \hat{k}$$

Because $\vec{a}$ is the angle bisector, it must be collinear with $\vec{d}$. This implies $\vec{a} = \lambda\vec{d}$ for some scalar $\lambda > 0$:

$$\alpha\hat{i} + 2\hat{j} + \beta\hat{k} = \lambda(\hat{i} + 2\hat{j} + \hat{k})$$

By equating the corresponding components of the vectors, we obtain the following system:

$$\alpha = \lambda$$

$$2 = 2\lambda \Rightarrow \lambda = 1$$

$$\beta = \lambda$$

### Final Calculation

Substituting the value $\lambda = 1$ into the equations for $\alpha$ and $\beta$, we find:

$$\alpha = 1$$

$$\beta = 1$$

The vector $\vec{a}$ is therefore $\hat{i} + 2\hat{j} + \hat{k}$. The values are **$\alpha = 1$ and $\beta = 1$**.

Select Answer:

Visualized Solution

Visualizing the Vectors

The Angle Bisector Concept

Magnitudes of $b$ and $c$

Finding the Bisector Direction

Relating $a$ to the Bisector

Setting up the Equation

Finding the Scalar $λ$

Finding $α$ and $β$

Final Conclusion

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Secret of Equal Magnitudes

The Algebraic Bridge

Final Calculation

Similar Questions

A vector $a = α \hat{i} + 2 \hat{j} + β \hat{k} (α, β \in R)$ lies in the plane of the vectors, $b = \hat{i} + \hat{j}$ and $c = \hat{i} - \hat{j} + 4 \hat{k}$ . If $a$ bisects the angle between $b$ and $c$ , then:

If the position vectors of the vertices A, B and C of a $Δ A B C$ are respectively $4 \hat{i} + 7 \hat{j} + 8 \hat{k}$ , $2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ , then the position vector of the point, where the bisector of $∠ A$ meets BC is :-

Let $a, b$ and $c$ be three non-zero vectors such that no two of these are collinear. If the vector $a + 2 b$ is collinear with $c$ and $b + 3 c$ is collinear with $a$ ( $λ$ being some non-zero scalar) then $a + 2 b + 6 c$ equals

If $C$ is the mid point of $A B$ and $P$ is any point outside $A B$ , then

Show, by vector methods, that the angular bisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices.

If $a$ and $b$ are unit vectors, then the greatest value of $3 ∣ a + b ∣ + ∣ a - b ∣$ is

If the vectors $A B = 3 \hat{i} + 4 \hat{k}$ and $A C = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle $A B C$ , then the length of the median through $A$ is

The vectors $A B = 3 \hat{i} + 4 \hat{k}$ and $A C = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle ABC. The length of the median through A is